Quantitative Aptitude PYQs

Concept:

(1 + x)n = nC0 × 1(n-0) × x 0+ nC1 × 1(n-1) × x 1 + nC2  × 1(n-2) × x2 + …. + nCn  × 1(n-n) × xn

n^th  term of the G.P. is an = arⁿ⁻¹

Sum of n terms = s = \(a (r^n-1)\over(r- 1)\); where r >1

Sum of n terms = s = \(a (1- r^n)\over(1- r)\); where r <1

Calculation:

C(n, 1) + C(n, 2) + _ _ _ _  _ + C(n, n) 

⇒ ⁿC₁ + ⁿC₂ + ... + ⁿC_n 

⇒ ⁿC₀ + nC1 + nC2 + ... + nCn - ⁿC₀

⇒ (1 + 1)ⁿ - ⁿC_o 

⇒ 2ⁿ - 1 = \(\rm 2^n - 1\over 2-1\) = 1 × \(\rm 2^n - 1\over 2-1\)

Comparing it with a G.P sum = a × \(\rm r^n - 1\over r-1\), we get a = 1 and r = 2

∴ 2n - 1 = 1 + 2 + 2² + ... +2^n-1 which will give us n terms in total.

Concept:

\(\rm ^n C_r = {n!\over(r!(n - r)!)}\)

(1 + x)ⁿ = ⁿC₀ × 1^(n-0) × x ⁰+ ⁿC₁ × 1^(n-1) × x ¹ + ⁿC₂  × 1(n-2) × x² + …. + ⁿC_n  × 1(n-n) × xⁿ

 

Calculation:

Given expansion is (1 + x)²ⁿ

⇒ ²ⁿC₀ ×1^(2n-0) × x⁰ +  2nC₁ ×1(2n-1) × x¹ + ... +  2nC_2n ×1(2n-2n) × x²ⁿ

First term = 2nC0 ×1 × 1 = 1

Last term =  2nC_2n ×1 × x²ⁿ = 1 × x²ⁿ = x²ⁿ

⇒ Sum = 1 + x²ⁿ

Coefficient of 1 = 1, coefficient of x²ⁿ = 1

∴ sum of the coefficients = 1 + 1 = 2.

Concept:

Let us consider sequence a₁, a₂, a₃ …. a_n is an A.P.

Common difference “d”= a₂ – a₁ = a₃ – a₂ = …. = a_n – a_{n – 1}

n^th term of the A.P.= a_n = a + (n – 1) d

Sum of the first n terms (S) =  \(\frac{n}{2}[2a+(n-1)d)]\)

Also, S =  (n/2)(a + l)

Where,

a = First term,

d = Common difference,

n = number of terms,

a_n = n^th term,

l = Last term

Calculation:

Given, a₁ = 2      ...(1)

S5 = \(1\over4\)(S10 - S5)

⇒ 4S5 = S10 - S5

⇒ 4S5 + S5 = S10

⇒ 5S5 = S10

⇒ 5 × \(5\over2\)[2a1 + (n5 - 1)d] = \(10 \over2\) [2a1 + (n10 - 1)d]

[∵ n5 = 5 और n10 = 10]

⇒ 5 × \(5\over2\)[a1 + a1 + 4d] = \(10 \over2\) [a1 + a1 + 9d]

⇒ 5 × [2a1 + 4d] = 2 × [2a1 + 9d]

⇒ 10a1 + 20d = 4a1 + 18d

⇒ 6a1 = 2d

⇒ a1 = \(\rm-d\over3\)

∴ d = -3a1 = - 3 × 2 = - 6

Hence,

S10 = \(10 \over2\) [a1 + a1 + 9d]

⇒ 5[2a1 + 9d]

⇒ 5[4 - 54] = -250

∴ S10 = 5 × (-50) = - 250.

Concept:

Multiplying or dividing with a non-zero number does not change the common ratio of the G.P

 

Calculation:

Multiplying or dividing with a non-zero number does not change the common ratio of the G.P

Let us consider sequence a1, a2, a3 …. an is a G.P.

Common ratio = r = \(\rm a_2\over a_1 \) = \(\rm a_3\over a_2 \) = \(\rm a_n\over a_{n-1} \)

If each term is multiplied/ divided by the same non-zero number it will be the same r value. So, the sequence will not change.

Concept:

Prime number: Prime number are those which are divisible by itself and 1.

Calculation:

Not a single five-digit prime number can be formed using the digits 1, 2, 3, 4, 5(without repetition).

This is because if one adds the digits, the result obtained will be = 1 + 2 + 3 + 4 + 5 = 15 which is divisible by 3.

Hence, any number obtained as a permutation of these 5 digits will be at least divisible by 3 and cannot be a prime number.

Concept:

To get the value of any function at any point with respect to some other function, we need to manipulate the given function/ point in such a way that we get the function that is needed.

 

Calculation:

Given, f(x + 1) = x2 - 3x + 2

f(x + 1) = x2 - 2x - x + 2......[-3x = -2x - x]

f(x + 1) = x(x - 2) - 1(x - 2)

f(x + 1) = (x - 2)(x - 1)

f(x + 1) = (x + 1 - 3)(x + 1 - 2).......[-2 = 1 - 3 & -1 = 1 - 2]

Now replace (x + 1) terms with (x)

∴ f(x) = (x - 3)(x - 2)

f(x) = x² -3x - 2x + 6

f(x) = x² - 5x + 6

Additional Information

This problem can also be solved by putting (x - 1) in place of x.

Concept:

Let us consider sequence a₁, a₂, a₃ …. an is an A.P.

Common difference “d”= a₂ – a₁ = a₃ – a₂ = …. = a_n – a_n-1

 

Calculation:

Since x², x, -8 are in A.P we can write,

x - x² = - 8 - x

⇒ x² - 2x - 8 = 0

⇒ x² - 4x + 2x - 8 = 0

⇒ x(x - 4) + 2(x - 4) = 0

⇒ (x - 4)(x + 2) = 0

⇒ x = -2, 4

∴ x ∈  {-2, 4}

Concept:

Let us consider sequence a₁, a₂, a₃ …. a_n is a G.P.

Common ratio = r = \(\rm a_2\over a_1 \) = \(\rm a_3\over a_2 \) = \(\rm a_n\over a_{n-1} \)

 

Calculation:

Consider,

(a = 3) be the 3^rd term of the G.P series,

So, we can write the five terms as,

\(\rm a\over r^2\), \(\rm a\over r\), a, ar, ar²

So, the product of the five terms (P) will be,

P = \(\rm a\over r^2\) × \(\rm a\over r\) × a × ar × ar2 = a⁵

Since,

a = 3,

∴ The product of the first five terms (P) = 3⁵ = 243

Concept:

If all the elements of a row or column are zeroes, then the value of the determinant is zero.

To evaluate the determinant row or column operation is done.

 

Calculation:

The determinant can be written as,

\(\begin{bmatrix} 2(1+1) & 2(1+2) &2(1+3) \ 2(2+1) &2(2+2) & 2(2+3)\ 2(3+1)&2(3+2) & 2(3+3) \end{bmatrix}\)

= \(2\times \begin{bmatrix} 1+1 &1+2 & 1+3\ 2+1 & 2+2 &2+3 \ 3+1 & 3+2 & 3+3 \end{bmatrix}\)

R₃ → R₃ - R₁ and R2 → R2 - R1

= \(2\times \begin{bmatrix} 1+1 &1+2 & 1+3\ 1 & 1 &1 \ 2 & 2 & 2 \end{bmatrix}\)

R₃ → R₃ -2R₂

= \(2\times \begin{bmatrix} 1+1 &1+2 & 1+3\ 1 & 1 &1 \ 0 & 0 & 0 \end{bmatrix}\)

= 0

So, the value of the determinant is 0

Concept:

Number of ways of arranging n object = n!

Calculation:

Given numbers are 2, 4, 6 & 8.

We can form determinant of order 2.

We know that the number of ways of arranging n object = n!

Number of determinants = 4! = 4 × 3 × 2 × 1 = 24

Let's form some determinants from the given numbers.

\(\begin{vmatrix} 2 & \rm 6 \ 8 & 4 \end{vmatrix}\) = 2 × 4 - 6 × 8 = 8 - 48 = - 40

\(\begin{vmatrix} 6 & \rm 2 \ 4 & 8 \end{vmatrix}\) = 48 - 8 = 40

\(\begin{vmatrix} 2 & \rm 8 \ 6 & 4 \end{vmatrix}\) = 8 - 48 = - 40

\(\begin{vmatrix} 6 & \rm 4 \ 2 & 8 \end{vmatrix}\) = 48 - 8 = 40

\(\begin{vmatrix} 4 & \rm 8 \ 6 & 2 \end{vmatrix}\) = 8 - 48 = - 40

\(\begin{vmatrix} 8 & \rm 4 \ 2 & 6 \end{vmatrix}\) = 48 - 8 = 40

\(\begin{vmatrix} 4 & \rm 6 \ 8 & 2 \end{vmatrix}\) = 8 - 48 = - 40

\(\begin{vmatrix} 8 & \rm 2 \ 4 & 6 \end{vmatrix}\) = 48 - 8 = - 40

Hence, we can see that we are getting the pattern where the other value will neutralize each determinant value.

Accordingly, the sum of the values of all determinants = 0

∴ The required sum is zero.

Concept:

The general form of the equation of a circle is:

x2 + y2 + 2gx + 2fy + c = 0

The centre of the circle is (-g, -f).

The radius of the circle is \(√{g^{2}+f^{2}-c}\).

Calculation:

Given equation of a circle is, 

4x2 + 4y2 - 20x + 12y - 15 = 0

x2 + y2 - 5x + 3y - 15/4 = 0

On comparing from the general equation of circle

2g = 5, or g = 5/2

2f = 3 or f = 3/2 & c = -15/4

By using the above formula

Radius = \(√{\frac{25}{4}+\frac{9}{4}-(-\frac{15}{4})}\)

Radius = \(√{\frac{25+9+15}{4}} =√ \frac{49}{4} \) = √12.25

Radius = 3.5

Alternate Method

Given equation of a circle is, 

4x2 + 4y2 - 20x + 12y - 15 = 0

⇒ [4x2 - 20x] + [4y2 + 12y] - 15 = 0

⇒ [(2x)² - 2 . 2x . 5 + 5² - 5²] + [(2y)2 + 2 . 2y . 3 + 32 - 32] - 15 = 0

⇒ [(2x - 5)2 - 52] + [(2y + 3)2 - 32] - 15 = 0

⇒ (2x - 5)2  + (2y + 3)2 - 15 - 25 - 9 = 0

⇒ (2x - 5)² + (2y +3)² = 49

⇒ 4(x - \(5\over2\))² + 4(y + \(3\over2\))² = 49

⇒ (x - \(5\over2\))2 + (y + \(3\over2\))² = (\(7\over2\))²

∴ The radius of the circle = \(7\over2\)units = 3.5 units

Concept:

ABCD is a parallelogram P, Q, R and S are mid-points of AB, BC, CD and DA respectively. Then PR and SQ bisect each other.

Calculation:

Applying the mid-point theorem,

\(\rm {{x+0}\over2}={{5-3}\over2}\)

⇒ x = 2

\({{y-4}\over2}={{4+2}\over2}\)

⇒ y = 10

So, the 4^th coordinate is (2, 10)

Concept:

Slope intercept form of a straight line ⇔ y = mx + c

Where m is the slope and c is the y-intercept

When two lines are perpendicular, the product of their slope is -1. If m is the slope of a line, then the slope of a line perpendicular to it is -1/m.

 

Calculation:

Given equations are, y + px = 1 and y - qx = 2 which can be written as,

y = - px + 1 and y = qx + 2 

So, m₁ (slope of 1^st line)= -p and m₂ (slope of 2^nd line)= q

As they are perpendicular m₁ × m₂ = - 1

⇒ - pq = - 1

⇒ pq - 1 = 0

Concept:

Let us consider sequence a1, a2, a3 …. an is an A.P.

Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an-1

 

Calculation:

Given A, B, C are in AP.

So we can write 2B = A + C

A - 2B + C = 0                           …(i)

Given equation of straight line is Ax + 2By + C = 0   …(ii)

Comparing (i) and (ii)

x = 1, y = -1

So (1, -1) lies on the line.

Concept:

For knowing the equation of the line mirror its slope needs to be determined first.

When two lines are perpendicular, the product of their slope is -1. If m is the slope of a line, then the slope of a line perpendicular to it is -1/m.

The slope of a line passing through the distinct points (x₁, y₁) and (x₂, y₂) is m =  \(\rm { (y_2 - y_1)}\over{(x_2 - x_1 )}\)

Equation of line is (y – y₁) = m(x – x₁)

 

Calculation:

The line mirror is perpendicular to the line joining the two points. So, for the point,

x = \({4-4}\over 2\) = 0

y = \({2-2}\over 2\) = 0

(x, y) = (0, 0)

Slope of the line joining (-4, 2) and (4, -2) = \({-2-2}\over{4-(-4)}\) = \(-1\over 2\)

As the mirror is perpendicular to this line so, m × (\(-1\over 2\)) = - 1

m = 2

Hence, the equation of line mirror is, \(\rm{y-0}\over{x-0}\) = 2

∴ y = 2x

Concept:

\(\rm tan^{-1}(x)+tan^{-1}({y})\) = \(\rm tan^{-1}\frac{x+y}{1-x\times y}\)

\(\rm cot^{-1}x=\) \(\rm tan^{-1}({1\over x})\)

Calculation:

Given, \(\rm tan^{-1}x+cot^{-1}x=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{1-x\times \frac{1}{x}} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}({\infty}) = \frac{\pi}{2}\)

This is true for all x ∈ R

Concept:

cos 2θ = \(\rm {(1 - tan^2\theta ) \over (1 + tan^2\theta)}\)

 

Calculation:

We know,

cos 2θ = \(\rm {(1 - tan^2\theta ) \over (1 + tan^2\theta)}\)

So, putting the value of tan A we can write,

cos 2A = \(\rm {(1 - tan^2A) \over (1 + tan^2A)}\)

∴ cos 2A  =  \(\rm \frac{1-\frac{1}{7}^2}{1+\frac{1}{7}^2}\)

∴ cos 2A  = \(\rm \frac{\frac{49-1}{7^2}}{\frac{49+1}{7^2}}\)

∴ cos 2A  = \(48 \over50\)

∴ cos 2A  = \(24\over25\)

Concept:

Cosine Rule of Triangle:

The square of the length of any side of a given triangle is equal to the sum of the squares of the length of the other sides minus twice the product of the other two sides multiplied by the cosine of angle included between them.

Consider, a, b, and c are lengths of the side of a triangle ABC as shown, then;

\(cos(x)=\frac{b^2+c^2-a^2}{2bc}\)

\(cos(y)=\frac{a^2+c^2-b^2}{2ac}\)

\(cos(z)=\frac{a^2+b^2-c^2}{2ab}\)

Calculation:

Let m = n = 1 unit

Then,

 \(\rm \sqrt{m^2+n^2+mn} \ \Rightarrow\rm \sqrt{1^2+1^2+1}= \sqrt 3\) 

Using cosine rule;

\(\rm \cos θ = {1^2+ 1^2 - {\sqrt 3}^2\over2\times 1 \times 1}\)

⇒ cos θ = -1/2

∴ θ = 120° 

Now, the sum of the acute angles of the triangle = 180° - 120° = 60°

Concept:

Area of a triangle = \(1\over 2\) × base × altitude

Area of ΔABC = \(\rm {1\over2} (a \cdot b\cdot sin \ C)\)

 

Calculation:

Area of a triangle = \(1\over 2\) × base × altitude

= \(1\over 2\) × c × a sin∠CBA

= \(1\over 2\) × 10 cm × 4cm sin 30° 

= 5 × 4 × \(1\over 2\) (as sin 30° = \(1\over 2\))

= 10 cm²

Concept:

Equivalent set: Two sets A and B are said to be equivalent if they have the same number of elements (cardinality), regardless of what the elements are. , i.e n(A) = n(B).

Equal set: Two sets A and B are said to be equal if they have exactly the same elements. The repetition of elements does not matter in sets because a set is defined as a collection of distinct objects, so repeated elements are generally ignored.

If A = B, then n(A) = n(B) and for any x ∈ A, x ∈ B too.

Calculation:

1. A = {1, 3, 5} and B = {2, 4, 7} are equivalent sets. → True

Set A has three elements: 1, 3, and 5. Set B also has three elements: 2, 4, and 7. Since both sets have three elements, they are indeed equivalent sets. 2. A = {1, 5, 9} and B = {1, 5, 5, 9, 9} are equal sets. → True

Set A contains the elements 1, 5, and 9. Set B also contains the elements 1, 5, and 9 (despite some elements being repeated, duplicates are not considered in set theory). Since both sets have the same elements, they are equal.

Therefore, both statements are correct:

Statement 1 is correct: A and B are equivalent sets.

Statement 2 is correct: A and B are equal sets.

Hence, The correct answer is "Both 1 and 2".

Source:- https://ncert.nic.in/textbook/pdf/kemh101.pdf, Page No. 7-8.